Problem 37 - Calculus Early Transcendentals

Question

Sketch the following regions \(R\). Then express \(\iint_{R} f(r, \theta) d A\) as an iterated integral over \(R\). The region outside the circle \(r=1\) and inside the rose \(r=2 \sin 3 \theta\) in the first quadrant

Step-by-Step Solution

Answer

Question: Determine the double integral representation in polar coordinates for a function \(f(r, \theta)\) over the region inside the polar curve \(r = 2\sin{3\theta}\) and outside the circle \(r = 1\) in the first quadrant. Answer: \(\iint_{R} f(r, \theta) dA = \int_{\theta_1}^{\theta_2} \int_1^{2\sin{3\theta}} f(r, \theta) rdrd\theta\), where \(\theta_1=\frac{1}{3}\sin^{-1}\left(\frac{1}{2}\right)\) and \(\theta_2=\frac{1}{3}\left(\pi-\sin^{-1}\left(\frac{1}{2}\right)\right)\).

Step 1: Sketch the regions

Let's begin by sketching the two given curves in polar coordinates. 1. The circle with \(r = 1\) is a simple circle centered at the origin with a radius of 1. 2. The rose with \(r = 2 \sin{3\theta}\) has petals with a length of 2, and there are 3 petals due to the multiple of 3 in the argument of sine. We can now sketch the two curves and shade the region inside the rose and outside the circle. We only consider the first quadrant, which is bounded by \(\theta=0\) and \(\theta=\frac{\pi}{2}\).

Step 2: Determine the bounds for \(r\) and \(\theta\)

To compute the double integral, we need to determine the bounds for \(r\) and \(\theta\). Note that, in the first quadrant: 1. The outer curve is the rose with \(r = 2 \sin{3\theta}\). 2. The inner curve is the circle with \(r = 1\). 3. The angles \(\theta\) are bounded by the first quadrant, i.e., \(0\leq \theta \leq \frac{\pi}{2}\). We need to find the points of intersection of the two curves to determine the bounds. To do this, set the equations of the two curves equal to each other and solve for \(\theta\): \(1 = 2 \sin{3\theta}\) \(\frac{1}{2} = \sin{3\theta}\) We can then find two angles for \(\theta\) (since we are in the first quadrant): \(\theta_1=\frac{1}{3}\sin^{-1}\left(\frac{1}{2}\right)\) and \(\theta_2=\frac{1}{3}\left(\pi-\sin^{-1}\left(\frac{1}{2}\right)\right)\)

Step 3: Express the double integral in polar coordinates

With the bounds for r and \(\theta\) determined, we can now express the double integral over the region \(R\) using polar coordinates. The double integral can be represented as: \(\iint_{R} f(r, \theta) dA = \int_{\theta_1}^{\theta_2} \int_1^{2\sin{3\theta}} f(r, \theta) rdrd\theta\) This is the final expression for the double integral over the region \(R\). The function \(f(r, \theta)\) can be replaced with any desired function to compute its integral over this region.

Key Concepts

Polar Coordinates

Polar coordinates provide a method of representing points in a plane using a radius and an angle. Unlike Cartesian coordinates, which use a grid of horizontal and vertical lines, polar coordinates describe a point by how far away it is from a fixed origin and the angle it makes with a reference direction.

Here's a brief overview:

• Radius (\(r\)): Indicates the distance from the origin, or center, of the coordinate system to the point.
• Angle (\(\theta\)): Measured counterclockwise from the positive x-axis to the line connecting the point with the origin.

Using these elements, any point in the plane can be expressed in terms of \(r\)and \(\theta\), making polar coordinates especially useful for dealing with curves and regions that have a circular symmetry, such as circles and roses, like those in the given problem.

In solving problems using polar coordinates, it's crucial to interpret equations correctly and understand how they relate to the geometric shapes they describe. This approach significantly simplifies the process when dealing with integrals over complex regions.

Iterated Integrals

Iterated integrals in the context of polar coordinates involve evaluating a double integral by integrating a function over a two-dimensional region described in terms of \(r\) and \(\theta\).

To perform an iterated integral:

• First, integrate with respect to \(r\), keeping \(\theta\) constant. This isolates a small ring or segment of the region.
• Next, integrate with respect to \(\theta\), considering the contribution of all these small segments, sweeping across the entire area.

The process of iterated integration is essential in transforming the area of interest into manageable pieces that can be summed up to give the total desired quantity.

In polar coordinates, the element of area is expressed as \(r dr d\theta\), so the double integral is typically represented as:\[\iint_{R} f(r, \theta) \, dA = \int_{a}^{b} \int_{c(\theta)}^{d(\theta)} f(r, \theta) \, r \, dr \, d\theta\]Where \([a, b]\) are the angular bounds and \([c(\theta), d(\theta)]\) are the radial bounds.

Regions in Polar Coordinates

Regions in polar coordinates are defined by radial and angular constraints, which describe boundaries within the polar plane. Understanding these regions aids in setting up iterated integrals and interpreting physical phenomena that exhibit circular symmetry.

For instance, in the exercise provided:

• The region is confined to the first quadrant, defined by \(0 \leq \theta \leq \frac{\pi}{2}\).
• Bounded outside by the rose curve \(r = 2 \sin 3 \theta\), and inside by the circle \(r = 1\).

To pinpoint the interaction points and set proper limits:

• Equate the boundaries of the inner and outer curves and solve for \(\theta\).
• The solutions will dictate the angular limits of integration. For example, where the rose meets the circle, as shown, provides the specific angles \(\theta_1\)and \(\theta_2\).

Mastering the interpretation of these regions allows for accurate integration and problem resolution, ensuring thorough comprehension of more complex systems where analytical precision is paramount.